3.132 \(\int x \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=38 \[ \frac{2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{4 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2} \]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (4*ArcTanh[Tanh[a + b*x]]^(9/2))/(63*b^2)

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Rubi [A]  time = 0.0142888, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac{2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{4 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (4*ArcTanh[Tanh[a + b*x]]^(9/2))/(63*b^2)

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac{2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{2 \int \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{7 b}\\ &=\frac{2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{2 \operatorname{Subst}\left (\int x^{7/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{7 b^2}\\ &=\frac{2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{4 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0598802, size = 32, normalized size = 0.84 \[ \frac{2 \left (9 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(2*(9*b*x - 2*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*b^2)

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Maple [A]  time = 0.033, size = 42, normalized size = 1.1 \begin{align*} 2\,{\frac{1/9\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}+1/7\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(tanh(b*x+a))^(5/2),x)

[Out]

2/b^2*(1/9*arctanh(tanh(b*x+a))^(9/2)+1/7*(b*x-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(7/2))

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Maxima [A]  time = 1.78357, size = 42, normalized size = 1.11 \begin{align*} \frac{2 \,{\left (7 \, b^{2} x^{2} + 5 \, a b x - 2 \, a^{2}\right )}{\left (b x + a\right )}^{\frac{5}{2}}}{63 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/63*(7*b^2*x^2 + 5*a*b*x - 2*a^2)*(b*x + a)^(5/2)/b^2

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Fricas [A]  time = 2.04863, size = 116, normalized size = 3.05 \begin{align*} \frac{2 \,{\left (7 \, b^{4} x^{4} + 19 \, a b^{3} x^{3} + 15 \, a^{2} b^{2} x^{2} + a^{3} b x - 2 \, a^{4}\right )} \sqrt{b x + a}}{63 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/63*(7*b^4*x^4 + 19*a*b^3*x^3 + 15*a^2*b^2*x^2 + a^3*b*x - 2*a^4)*sqrt(b*x + a)/b^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.11733, size = 178, normalized size = 4.68 \begin{align*} \frac{\sqrt{2}{\left (\frac{21 \, \sqrt{2}{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} a^{2}}{b} + \frac{6 \, \sqrt{2}{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} a}{b} + \frac{\sqrt{2}{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )}}{b}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/315*sqrt(2)*(21*sqrt(2)*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)*a^2/b + 6*sqrt(2)*(15*(b*x + a)^(7/2) - 42
*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*a/b + sqrt(2)*(35*(b*x + a)^(9/2) - 135*(b*x + a)^(7/2)*a + 189*(
b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)/b)/b